Let $\psi$ be a mollifier with $spt(\psi) \subset B_1(0)$. Define $\psi_\epsilon$ as usual. Now let $\mu$ be a finite borel measure. Define $\mu_\epsilon = \psi_\epsilon * \mu$. Then $\hat{\mu_\epsilon}$ converges uniformly to $\hat{\mu}$.
But how to show the statement? We know that $\mu_\epsilon$ converges weakly to $\mu$ but how does this help us?
Any hints/answer would be appreciated.
This is false, assuming the domain is the real line. Let $\mu$ be the Dirac delta at the origin. Then $\widehat{\mu}=1$ and $\widehat{\mu}_\epsilon=\widehat{\phi}_\epsilon$ is a function that decays to $0$ at infinity for all $\epsilon$.
To make this true, you would need an additional assumption that $\widehat{\mu}$ also has some decay. Then
$\widehat{\mu}_\epsilon-\widehat{\mu}=(\widehat{\phi}_\epsilon-1)\widehat{\mu}$, and $\widehat{\phi}_\epsilon$ is converging to 1 uniformly on all compact subsets.